$n_{Zn}=16,25/65=0,25mol$
$PTHH : Zn+2HCl\to ZnCl_2+H_2$
a.Theo pt :
$n_{HCl}=2.n_{Zn}=0,25.2=0,5mol$
$⇒A_{HCl}=0,5.6.10^{23}=3.10^{23}$
b.Theo pt :
$n_{ZnCl_2}=n_{Zn}=0,25mol$
$⇒m_{ZnCl_2}=0,25.136=34g$
Tên muối : Kẽm clorua
$c.2Zn+O_2\overset{t^o}\to 2ZnO$
Theo pt :
$n_{O_2}=1/2.n_{Zn}=1/2.0,25=0,125mol$
$⇒V_{kk}=0,125.22,4.5=14l$